# Compounding

Before getting to the general rules I’ve come up with lets take a quick look at the most famous method to solve compounding questions.

### The rule of 72: what is it and why does it work

`"In order to find the time it takes to double your money divide 72 by the rate of return"`

For example if your rate of return is 8% it should take you 72/8=9 years to double. The actual answer is 9.006 so not bad at all. Let’s take a quick look at why the rule works. I will basically repeat what was done here if you want to see the original. The rule is basically trying to solve 2x=x(1+r)y .

`Rearrange for y=ln(2)/ln(1+r) We are looking for something in the form y=k/r Equate k/r=ln(2)/ln(1+r) to get k=r*ln(2)/ln(1+r) where we use rule of k`

This is where the 8% part comes in, if we substitute ‘r’ for 0.08 we get k=0.7205. Times this by 100 so we can use 8 instead of 0.08 and we get 72.05 or roughly 72. Now if we change ln(2) to ln(y) depending on what value magnitude we have we get (centred around 8%)

### What’s wrong

There are a few issues to this rule in my view, they are that:

• It is most accurate for 8% and values near there. We will see why in a bit but basically if you want to correct the formula you need to add 1 to 72 for every 4% above 8% and subtract 1 for every 4% below 8%. For example if you had a rate of return of 12% you should use 72+1=73 for your rule. This is especially important if you are looking at returns in the 20% kind of range e.g Berkshire Hathaway returns.
• It is a rule for doubling, this means it can be used if you quadruple or make 8 times you money but for intermediate values it can be tricky to use. We can fix this problem however if we use different rules for different multiples, for example if we want to find the time to triple your money we can use a rule of 114 but learning these values for all different multiples can be tricky and you will see the issues if you try to find the time to increase your money by 1000 times for instance.

### So can we generalise?

There are three cases we want to be able to solve for. If we are given a:

1. Magnitude increase and interest rate and want to know the number of years
2. Magnitude increase and number of years and want to know the interest rate
3. Number of years and interest rate and want to know the magnitude increase

Before having a look at these cases it’s worth recapping somethings I’ve posted about before. To use any of the rules below I assume that you can work out any natural logarithm. This may seem like a big ask but because of how logarithms work if you know a few, I recommend the values for the primes, then you can figure out any logarithm (more about this here).

In all cases I have used r to mean the rate of return, y as the magnitude of increase, and n as the number of years.

#### Case 1: Magnitude and interest rate

The rules above work best around interest rates of 8% and even then you would still need to learn each rule. Let’s try to simplify things. Take k=[r/ln(1+r)]ln(2), (we saw this formula at the beginning), if we plot r/ln(1+r) for percentages up to 25% for instance we get an almost linear line and a rule seems to form, that is for return r% we divide it by 2 and add it to 1. For example for 20% we have 1+(0.2/2)=1.1

### Rule: n = [(1+r/2)ln(y)*100 ]/(r*100)

This may look unfriendly but it’s basically just increasing some number by a certain percentage and then dividing it by something else. [note for the r/2 term, we use the decimal version of percentages not the number form, e.g for 8% we use 0.08 rather than 8]. We now have a rule that will adjust for different percentages, and if you can work out different logarithms we can adjust for different magnitudes.

Example: How long does it take to increase your money by 3 times at return of 20%?

• We get (1+(0.2/2))*ln(3)*100 = 1.1*ln(3)*100
• We know that ln(3) is approximately 1.1 so 100*1.1 is 110
• We therefore want about 10% above this so we want a rule of 121
• Divide 121 by 20 (this is the r*100 bit) and we get 6.05 years (the actual answer is 6.025 years so not bad)

#### Case 2: Magnitude and number of years

This is similar to the above apart from the adjustment is a bit different. Here we increase our logarithm by 7% for all values. For example

### Rule: r = [1.07 * ln(y)*100]/n

Example: What interest rate do you need to increase your money 5 times in 20 years?

• ln(5) = 1.6
• 1.07*1.6*100 = 171.2 so we get a rule of roughly 171
• 171/20 = 8.5% (the exact answer is 8.38 so not bad)

One thing to be aware of is that if the interest rate that comes out is above 20% then the answer from this is likely an underestimation. If you are in the low 20%’s then you will probably want to ad 0.5% to your answer. If you are in high 20%’s then probably add around 1% to 1.5%.

#### Case 3: Number of years and interest rate

This technique is very different from the others and probably the most challenging. You end up having to figure out how to make a number just by adding things like ln(2), ln(3) etc together which can be a bit harder. This means we don’t have a rule so much as a process to follow.

### Process:x = n*r*100x*0.95 (reduce answer by 5%)z = x/100find how to make z using natural logarithmstake the natural logarithms you used and times them together

Example: How much does your money increase by if you make 7% for 20 years?

• 7*20 = 140
• 140*0.95 = 132
• 132/100 = 1.32
• 1.32 is close to 1.4 which is ln(2)+ln(2)
• Since we used ln(2) twice we get an answer of 2*2 = 4
• (in this case as 1.32 was a bit below 1.4 we would guess a bit below 4 say 3.8)
• The actual answer is 3.87

Example 2: How much does your money increase by if you make 6% for 50 years?

• 6*50 = 300
• 300*0.95 = 285
• 285/100 = 2.85 (close to 2.9)
• ln(3)+ln(3)+ln(2) = 1.1+1.1+0.7 = 2.9